lecture

1. Hydrostatic Force on a Curved Surfaces
Henryk Kudela
1 Hydrostatic Force on a Curved Surface
On a curved surface the forces pδA on individual elements differ in direction, so a simple
summation of them may not be made. Instead, the resultant forces in certain directions
may be determined, and these forces may then be combined vectorially. It is simplest to
calculate horizontal and vertical components of the total force.
Horizontal component of hydrostatic force
Any curved surface may be projected on to a vertical plane. Take, for example, the curved
surface illustrated in Fig. 1.
Figure 1: Hydrostatic force on a curved surface
Its projection on to the vertical plane shown is represented by the trace AC. Let Fx
represent the component in this direction of the total force exerted by the fluid on the
curved surface. Fx must act through the center of pressure of the vertical projection and
is equal in magnitude to the force F on the fluid at the vertical plane.
In any given direction, therefore, the horizontal force on any surface equals the force
on the projection of that surface on a vertical plane perpendicular to the given direction.
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2. The line of action of the horizontal force on the curved surface is the same as that of the
force on the vertical projection.
Vertical component of hydrostatic force
The vertical component of the force on a curved surface may be determined by considering
the fluid enclosed by the curved surface and vertical projection lines extending to the free
surface. Thus
FH = F2 = ρgzs (1)
FV = F1 = ρgV (2)
where V is the volume of the liquid between the free surface liquid and solid curved surface.
The magnitude of the resultant is obtained form the equation
FR = F2
H + F2
V (3)
Example 1 A sector gate, of radius 4 m and length 5m, controls the flow of water in a
horizontal channel. For the (equilibrium) conditions shown in Fig. 2 , determine the total
thrust on the gate.
Figure 2: Hydrostatic force at a curved gate
Solution.
Since the curved surface of the gate is part of a cylinder, the water exerts no thrust along its
length, so we consider the horizontal and vertical components in the plane of the diagram.
The horizontal component is the thrust that would be exerted by the water on a vertical
projection of the curved surface. The depth d of this projection is 4 sin 30o = 2 m, (R =
4 m) and its centroid is 1 + d/2 = 2 m below the free surface. Therefore horizontal force
FH is equal
FH = ρgzsA = 1000 · 9.81 · 2 · (5 · 2) = 1.962 · 105
N
Its line of action passes through the center of pressure of the vertical projection, that is,
at a distance Ix/Azs below the free surface, given by:
Ix
Azs
=
Is + Az2
s
Azs
=
bd3
12(b d)zs
+ zs =
2
12 · 2
+ 2 + 1.167 m
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3. The vertical component of the total thrust is equal of weight of imaginary water ABC.
AB = (4 − 4 cos 30o
) = 0.536 m
Vertical force is equal FV = ρgV . Then
FV = 1000 · 9.81 · 5 (0.536 · 1) + π · 42
·
30
360
−
1
2
· 2 · 4 cos 30o
= 6.18 · 104
N
The horizontal and vertical components are co-planar and therefore combine to give a
single resultant force of magnitude
|F| = F2
H + F2
V = (1.962105
)2
+ (6.18104)2
= 2.057 · 105
N
at angle equal to arctg( 6180
196200 ) ∼= 17.5o to the horizontal axis.
Problem 1.
The dam in Fig. 3 is a quarter circle 50 m wide into the paper. Determine the horizontal
and vertical components of the hydrostatic force against the dam and the point CP where
the resultant strikes the dam.
Figure 3: Hydrostatic force at a curved dam
(Ans. FH = 97.9 MN, FV = 153.8 MN)
Problem 2.
Compute the horizontal and vertical components of the hydrostatic force on the quarter-
circle panel at the bottom of the water tank in Fig. 4
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4. Figure 4: Hydrostatic force at the bottom of water tank
2 Buoyancy
Because the pressure in a fluid in equilibrium increases with depth, the fluid exerts a
resultant upward force on any body wholly or partly immersed in it. This force is known
as the buoyancy. This buoyancy force can be computed using the same principles used
to compute hydrostatic forces on surfaces. The results are the two laws of buoyancy
discovered by Archimedes in the third century B.C.:
1. a body immersed in a fluid experiences a vertical buoyant force equal to the weight
of the fluid it displaces.
2. a floating body displaces its own weight in the fluid in which it floats.
So
FB = Vbodyρg (4)
In fig. 5 the vertical force exerted on an element of the body in the form of vertical prism
of cross section δA is
δF = (p2 − p1)δA = ρghδA = ρgδV
in which δV is the volume of the prism. Integrating over the complete body gives the
formula (4).
Weighting an odd-shaped object suspended in two different fluids yields sufficient data
to determine its weight W, volume V, specific weightγ = ρg (or density ρ) and specific
gravity SG = ρ
ρwater
. Figure 6 shows two free-body diagrams for the same object suspended
and weighed in two fluids. F1 and F2 are the weight when submerged and γ1 = ρ1g and
γ2 = ρ2g are the specific weights of the fluids. W and V , the weight and volume of the
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5. Figure 5: Vertical force componets on element of body
object, respectively, are to be found. The equations of equilibrium are written
F1 + V ρ1g = W, F2 + V ρ2g = W
and solved
V =
F1 − F2
g(ρ1 − ρ2)
, W =
F1ρ1 − F2ρ2
ρ2 − ρ1
The hydrometr uses the principle of buoyant force to determine specific gravities of liquids.
Figure 6: Free-body diagrams for body suspended in a fluid
Figure 7 shows a hydrometr in two liquids. It has a stem of prismatic cross section a.
Considering the liquid on the left to be distilled water, SG = 1.00, the hydrometr floats in
equilibrium when
V0ρg = W
in which V0 is the volume submerged, ρg = γ is the specific weight of water, and W is the
weight of the hydrometr. The position of the liquid surface is marked as 1.00 on the stem
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6. to indicate unit specific gravity SG. When the hydrometr is floated in another liquid the
equation of equilibrium becomes
(V0 − ∆V )SGρg = W
in which ∆V = a∆h. Solving for ∆h using above equations gives
∆h =
V0
a
SG − 1
SG
from which the stem can be marked off to read specific gravities or densities.
Figure 7: Hydrometr in water and in liquid of specific graviry SG.
References
[1] F. M. White, 1999. Fluid Mechanics, McGraw-Hill.
[2] B. R. Munson, D.F Young and T. H. Okiisshi, 1998. Fundamentals of Fluid Mechanics,
John Wiley and Sons, Inc. .
[3] B. S. Massey and J. Ward-Smith, 2006, Mechanics of Fkuids, Taylor and Francis
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